Solutions

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Congruences from A Study Guide for Beginner’s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair

29. Solve the congruence

42x ≡ 12 (mod 90).

Comment: You need to recall Theorem 1.3.5, which states that ax ≡ b (mod n) has a solution if an only if gcd(a, n) is a divisor of b. Also note that the congruence is stated modulo 90, and so the most satisfying answer is given in terms of congruence classes modulo 90. Solution: We have gcd(42, 90) = 6, so there is a solution since 6 is a factor of 12. Solving the congruence 42x ≡ 12 (mod 90) is equivalent to solving the equation 42x = 12+ 90q for integers x and q. This reduces to 7x = 2+ 15q, or 7x ≡ 2 (mod 15). (Equivalently, we could obtain 7x ≡ 2 (mod 15) by dividing 42x ≡ 12 (mod 90) through by 6.) We next use trial and error to look for the multiplicative inverse of 7 modulo 15. The numbers congruent to 1 modulo 15 are 16, 31, 46, 61, etc., and −14, −29, −44, etc. Among these, we see that 7 is a factor of −14, so we multiply both sides of the congruence by −2 since (−2)(7) = −14 ≡ 1 (mod 15). Thus we have −14x ≡ −4 (mod 15), or x ≡ 11 (mod 15). The solution is x ≡ 11, 26, 41, 56, 71, 86 (mod 90). 30. (a) Find all solutions to the congruence

55x ≡ 35 (mod 75).

Solution: We have gcd(55, 75) = 5, which is a divisor of 35. Thus we have 55x ≡ 35 (mod 75);

11x ≡ 7 (mod 15);

−x ≡ 13 (mod 15);

x ≡ 2 (mod 15).

44x ≡ 28 (mod 15); The solution is

x ≡ 2, 17, 32, 47, 62 (mod 75). Comment: In the solution, the congruence 11x ≡ 7 (mod 15) is multiplied by 4 since trial and error produces the congruence 4 · 11 ≡ −1 (mod 15), a relatively easy way to eliminate the coefficient of x. (b) Find all solutions to the congruence

55x ≡ 36 (mod 75).

Solution: There is no solution, since gcd(55, 75) = 5 is not a divisor of 36. 31. (a) Find one particular integer solution to the equation 110x + 75y = 45. Solution: By Theorem 1.1.6, any linear combination of 110 and 75 is a multiple of their greatest common divisor. We have following matrix reduction. 1 0 110 1 −1 35 1 −1 35 15 −22 0 ; ; ; 0 1 75 0 1 75 −2 3 5 −2 3 5 Thus −2(110) + 3(75) = 5, and multiplying by 9 yields a solution: x = −18, y = 27, since 110(−18) + 75(27) = 45. Alternate solution: The equation reduces to the congruence 35x ≡ 45 (mod 75). This simplifies to 7x ≡ 9 (mod 15), and multiplying both sides by −2 gives x ≡ −3 (mod 15). Thus 75y = 45 + 3(110) = 375 and so x = −3, y = 5 is a solution.

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Comment: The matrix computation (above) shows that 110(15) + 75(−22) = 0, so adding any multiple of the vector (15, −22) to the particular solution (−18, 27) must also give you a solution. That is the motivation for part (b) of the problem. (b) Show that if x = m and y = n is an integer solution to the equation in part (a), then so is x = m + 15q and y = n − 22q, for any integer q. Solution: If 110m + 75n = 45, then 110(m + 15q) + 75(n − 22q) = 45 + 110(15)q + 75(−22)q = 45, since 110(15) − 75(22) = 0. 32. Solve the system of congruences

x ≡ 2 (mod 9)

x ≡ 4 (mod 10) .

Solution: We can easily find a linear combination of 9 and 10 that equals 1, by just writing (1)(10) + (−1)(9) = 1. Using the method outlined in the proof of Theorem 1.3.6, the solution is x ≡ (2)(1)(10) + (4)(−1)(9) = −16 (mod 90). Alternate solution: Convert the second congruence to the equation x = 4 + 10q for some q ∈ Z, and substitute for x in the second congruence. Then 4+ 10q ≡ 2 (mod 9), which reduces to q ≡ 7 (mod 9). The solution is x ≡ 4 + 10(7) ≡ 74 (mod 90). 33. Solve the system of congruences

x ≡ 5 (mod 25)

x ≡ 23 (mod 32) . 1 0 32 ; Solution: To solve r(32) + s(25) = 1 we will use the matrix method. 0 1 25 1 −1 7 1 −1 7 4 −5 3 4 −5 3 ; ; ; Thus 0 1 25 −3 4 4 −3 4 4 −7 9 1 (−7)(32) + (9)(25) = 1, and so x ≡ (5)(−7)(32) + (23)(9)(25) = 4025 ≡ 55 (mod 800). Alternate solution: Write x = 23+32q for some q ∈ Z, and substitute to get 23+32q ≡ 5 (mod 25), which reduces to 7q ≡ 7 (mod 25), so q ≡ 1 (mod 25). This gives x ≡ 55 (mod 800).

34. Solve the system of congruences

5x ≡ 14 (mod 17)

3x ≡ 2 (mod 13) .

Solution: By trial and error, 7 · 5 ≡ 1 (mod 17) and 9 · 3 ≡ 1 (mod 13), so 5x ≡ 14 (mod 17);

35x ≡ 98 (mod 17);

x ≡ 13 (mod 17)

and 3x ≡ 2 (mod 13);

27x ≡ 18 (mod 13);

x ≡ 5 (mod 13).

Having reduced the system to the standard form, we can solve it in the usual way. We have x = 13 + 17q for some q ∈ Z, and then 13 + 17q ≡ 5 (mod 13). This reduces to 4q ≡ 5 (mod 13), so 40q ≡ 50 (mod 13), or q ≡ 11 (mod 13). This leads to the answer, x ≡ 13 + 17 · 11 ≡ 200 (mod 221). 35. Give integers a, b, m, n to provide an example of a system x ≡ a (mod m)

x ≡ b (mod n)

that has no solution. Solution: In the example the integers m and n cannot be relatively prime. This is the clue to take m = n = 2, with a = 1 and b = 0.

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36. Find the additive order of each of the following integers, module 20: 4, 5, 6, 7, and 8. Note: The additive order of a modulo n is defined to be the smallest positive solution of the congruence ax ≡ 0 (mod n). Solution: To find the additive order of 4, we need to solve the congruence 4x ≡ 0 (mod 20). Dividing each term by gcd(4, 20) = 4, we obtain x ≡ 0 (mod 5), and then the smallest positive solution is x = 5. Thus 4 has additive order 5 modulo 20. The additive order of 5 modulo 20 is 4, as shown by this solution of 4x ≡ 0 (mod 20). 5x ≡ 0 (mod 20)

x ≡ 0 (mod 4)

x = 4.

The additive order of 6 modulo 20 is 10: 6x ≡ 0 (mod 20)

3x ≡ 0 (mod 10)

x ≡ 0 (mod 10)

x = 10.


x ≡ 0 (mod 20)

x = 20.


2x ≡ 0 (mod 5)

x ≡ 0 (mod 5)

x = 5.

37. (a) Compute the last digit in the decimal expansion of 4100 . Solution: The last digit is the remainder when divided by 10. Thus we must compute the congruence class of 4100 (mod 10). We have 42 ≡ 6 (mod 10), and then 62 ≡ 6 (mod 10). This shows that 4100 = (42 )50 ≡ 650 ≡ 6 (mod 10), so the units digit of 4100 is 6. (b) Is 4100 divisible by 3? Solution: No, since 4100 ≡ 1100 ≡ 1 (mod 3). Or you can write 2200 as the prime factorization, and then gcd(3, 2200 ) = 1. 38. Find all integers n for which 13 | 4(n2 + 1). Solution: This is equivalent to solving the congruence 4(n2 + 1) ≡ 0 (mod 13). Since gcd(4, 13) = 1, we can cancel 4, to get n2 ≡ −1 (mod 13). Just computing the squares modulo 13 gives us (±1)2 = 1, (±2)2 = 4, (±3)2 = 9, (±4)2 ≡ 3 (mod 13), (±5)2 ≡ −1 (mod 13), and (±6)2 ≡ −3 (mod 13). We have done the computation for representatives of each congruence class, so the answer to the original question is n ≡ ±5 (mod 13). Comment: For example, if n = 5, then 13 | 4 · 26. 39. Prove that 10n+1 + 4 · 10n + 4 is divisible by 9, for all positive integers n. Comment: This could be proved by induction, but we can give a more elegant proof using congruences. Solution: The proof consists of simply observing that 10n+1 + 4 · 10n + 4 ≡ 0 (mod 9) since 10 ≡ 1 (mod 9). 40. Prove that for any integer n, the number n3 + 5n is divisible by 6. Solution: By Proposition 1.2.3 (c), it is enough to show that n3 + 5n ≡ 0 (mod 2) and n3 + 5n ≡ 0 (mod 3), reducing the question to just a few computations. Modulo

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2, we have 03 + 5(0) ≡ 0 (mod 2), and 13 + 5(1) = 6 ≡ 0 (mod 2). Modulo 3, we have 03 + 5(0) ≡ 0 (mod 3), 13 + 5(1) = 6 ≡ 0 (mod 3), and 23 + 5(2) ≡ 8 + 10 ≡ 0 (mod 3). Therefore 6 | n3 + 5n. 41. Use techniques of this section to prove that if m and n are odd integers, then m2 − n2 is divisible by 8. (Compare Problem 1.2.36.) Solution: We need to show that if m and n are odd, then m2 − n2 ≡ 0 (mod 8). Modulo 8, any odd integer is congruent to either ±1 or ±3, and squaring any of these four values gives 1 (mod 8). Thus m2 − n2 ≡ 1 − 1 ≡ 0 (mod 8). 42. Prove that 42n+1 − 74n−2 is divisible by 15, for all positive integers n. Solution: We have 42 ≡ 1 (mod 15), so 42n+1 = (42 )n · 4 ≡ 4 (mod 15). We also have 72 ≡ 4 (mod 15), so 74 ≡ 1 (mod 15), and thus 74n−2 ≡ 72 · (74 )n−1 ≡ 4 (mod 15). Therefore 42n+1 − 74n−2 ≡ 4 − 4 ≡ 0 (mod 15). Alternate solution: By Proposition 1.2.3 (c), it is enough to show that 42n+1 −74n−2 ≡ 0 (mod 3) and 42n+1 − 74n−2 ≡ 0 (mod 5). We have 42n+1 − 74n−2 ≡ 12n+1 − 14n−2 ≡ 1 − 1 ≡ 0 (mod 3) and 42n+1 − 74n−2 ≡ (−1)2n+1 − 22(2n−1) ≡ (−1)2n+1 − (22 )2n−1 ≡ (−1)2n+1 − (−1)2n−1 ≡ −1 − (−1) ≡ 0 (mod 5). 43. Prove that the fourth power of an integer can only have 0, 1, 5, or 6 as its units digit. Solution: Since the question deals with the units digit of n4 , it is really asking to find n4 (mod 10). All we need to do is to compute the fourth power of each congruence class modulo 10: 04 = 0, (±1)4 = 1, (±2)4 = 16 ≡ 6 (mod 10), (±3)4 = 81 ≡ 1 (mod 10), (±4)4 ≡ 62 ≡ 6 (mod 10), and 54 ≡ 52 ≡ 5 (mod 10). This shows that the only possible units digits for n4 are 0, 1, 5, and 6. ANSWERS AND HINTS 45. Solve the following congruences. (a) 10x ≡ 5 (mod 21) Answer: (b) 10x ≡ 5 (mod 15) Answer: (c) 10x ≡ 4 (mod 15) Answer: (d) 10x ≡ 4 (mod 14) Answer:

x ≡ 11 (mod 21) x ≡ 2, 5, 8, 11, 14 (mod 15) No solution x ≡ 6, 13 (mod 14)

47. Solve the following congruence. 20x ≡ 12 (mod 72) Answer: x ≡ 15, 33, 51, 69 (mod 72) 49. (a) Find the additive order of each of the following elements, by solving the appropriate congruences. 4, 5, 6 modulo 24 Answer: The congruence class of 4 has additive order 6, that of 5 has additive order 24, and that of 6 has additive order 4 (modulo 24). 53. Solve the following system of congruences: Answer: x ≡ 43 (mod 400)

x ≡ 11 (mod 16)

55. Solve the following system of congruences: Answer: x ≡ −41 ≡ 409 (mod 450)

x ≡ 9 (mod 25)

x ≡ 18 (mod 25) x ≡ 13 (mod 18)

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57. Solve this system: 2x ≡ 3 (mod 7) Answer: x ≡ 208 (mod 462)

x ≡ 4 (mod 6)

5 5x ≡ 50 (mod 55)

59. Use congruences to prove that 52n − 1 is divisible by 24, for all positive integers n. Solution: We have 52n = (52 )n ≡ 1n ≡ 1 (mod 24). 61. Prove that if 0 < n < m, then 22 + 1 and 22 + 1 are relatively prime. n m Hint: Write 22 as a power of 22 . If p is a common prime divisor, reduce modulo p. n

m